# Finding $x$ from inequality: $\left | \frac{3^n + 2}{3^n + 1} - 1 \right | \le \frac{1}{28}$

Find $x$ in $\mathbb{Z}$ satisfying this inequality:

$$\left | \frac{3^n + 2}{3^n + 1} - 1 \right | \le \frac{1}{28}.$$

I tried something, but I don't think it's correct.

$$-\frac{1}{28} \le \frac{3^n + 2}{3^n + 1} - 1 \le \frac{1}{28}$$

I arrived at:

$$-3^n - 1 \le 28 \le 3^n + 1.$$

I don't know if it's ok or not.

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It might help you to combine $${3^n+2\over3^n+1}-1$$ into a single fraction. – Gerry Myerson Aug 4 '13 at 11:38

Hints:

$$\left|\;\frac{3^n+2}{3^n+1}-1\;\right|=\frac1{3^n+1}\le\frac1{28}\iff\ldots$$

Note that $\,3^n+1>0\;,\;\;\forall\,n\in\Bbb N\;$

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I have nothing more that you noted ;) Just to make a machinery approach. That's it. Of course the OP didn't want it:) – Babak S. Aug 4 '13 at 12:59
i arrived here : -1/28<=1/(3^n+1)<=1/28 and i don't know what else to do.I tried to reduce to a common denominator and i got this: −3n−1≤28≤3n+1 Can you tell me the next step? – marinaaaa Aug 4 '13 at 13:45
@marinaaaa, you don't need all that: the expression inside the absolute value is always positive! You can erase the absolute value and work without it, as shown above... – DonAntonio Aug 4 '13 at 14:12

the answer is [3, infinity) because n is greater than or = 3

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This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. – Dan Rust Aug 4 '13 at 12:46